Problem 10 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{array}{l} 2 x+3 y+7 z=13 \\ 3 x+2 y-5 z=-22 \\ 5 x+7 y-3 z=-28 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the system of equations is \(x = -204, y = 145, z = -2\).

Step 1: Eliminating a Variable

Start by eliminating one of the variables in any two of the equations. Here, choose the first and second equation and eliminate the variable x. Multiply the first equation by 3 and the second equation by 2 and then subtract the second from the first: \(6x + 9y + 21z = 39\) and \(6x + 4y - 10z = -44\). The resulting equation is \(5y + 31z = 83\). Let's call this Equation 4.

Step 2: Further Elimination of the Same Variable

Perform the elimination step again, but this time with the first and third equation. Multiply the first equation by 5 and the third equation by 2 and then subtract the third from the first: \(10x + 15y + 35z = 65\) and \(10x + 14y - 6z = -56\). The resulting equation is \(y + 41z = 121\). Let's call this Equation 5.

Step 3: Solve the New System of Equations from Step 1 and Step 2

From Equation 4, isolate y to get \(y = 83 - 31z\). Substituting y from Equation 4 into Equation 5 yields \(83 - 31z + 41z = 121\), simplifying gives \(z = -2\).

Step 4: Finding y and x

Substitute \(z = -2\) into Equation 4 to find the value of y: \(y = 83 - 31*(-2) = 145\). Similarly, substitute \(z = -2\) and \(y = 145\) into the first original equation to find the value of x: \(2x + 3*145 + 7*(-2) = 13, x = -204\)